# Thread: Hyperbolic Function & e

1. ## Hyperbolic Function & e

Q: Starting from the definition of $\mathrm{cosh}x$ in terms of $e^x$, find, in terms of natural logarithms, the value for $x$ for which $5 = 3 \mathrm{cosh}x$.

I did:
$5 = 3 \mathrm{cosh}x$
$5 = 3 \left( \frac{e^x + e^{-x}}{2} \right)$
$10 = 3e^x + 3e^{-x}$
$0 = 3e^x + e^{-x} - 10$
$0 = 3e^{2x} - 10e^x + 1$

However, that does not factorise. The answer is $\pm \ln 3$. Can someone do this question to see if they can get the answer. If possible, please show me your steps. Thanks in advance.

2. Originally Posted by Air
I did:
$5 = 3 \mathrm{cosh}x$
$5 = 3 \left( \frac{e^x + e^{-x}}{2} \right)$
$10 = 3e^x + 3e^{-x}$
$0 = 3e^x + {\color{red}3}e^{-x} - 10$
Where did the 3 go?

3. ^ Thanks!

Also, is $\ln \left( \frac{1}{3} \right) = - \ln (3)$?

4. Yep: $\ln \left(\frac{1}{3}\right) = \ln \left(3^{-1}\right) = -\ln 3$

where: $\ln (a)^{n} = n\ln a$

5. Another follow on question is: Find the minimum value of $(5 \mathrm{cosh}x + 3 \mathrm{sinh}x$.

I have no idea how to do this. Can someone help me on how to do this. Thanks.

6. If you recall what you've learned from derivatives, you know that your local minimum can be found by finding f'(x) = 0 and verifying with your first derivatives test. So:

$f(x) = 5\left(\frac{e^{x} + e^{-x}}{2}\right) + 3\left(\frac{e^{x} - e^{-x}}{2}\right)$
$f(x) = \frac{5e^{x} + 5e^{-x} + 3e^{x} - 3e^{-x}}{2}$

Simplify and find for what values f'(x) = 0.